Amines Test

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Amines Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The correct IUPAC name for $$CH_{2}=CHCH_{2}NHCH_{3}$$ is

A
Allylmethylemine

B
$$2-$$ amino $$-4-$$ pentene

C
$$4-$$ aminopent $$-1-$$ ene

D
$$N-$$ methylprop $$-2-$$ en $$-1-$$ amine

Solution

$$IUPAC$$ name of the compound $$CH_2=CHCH_2NHCH_3$$ is $$N-$$ methyprop $$-2-$$ en $$-1-$$ amine.
Question 2
1 / -0

The correct decreasing order of basic strength of the following species is:

$$H_{2}O, NH_{3}, OH^{-}, NH_{2}^{-}$$

A
$$NH_{2}^{-} > OH^{-} > NH_{3} > H_{2}O$$

B
$$OH^{-} > NH_{2}^{-} > H_{2}O > NH_{3}$$

C
$$NH_{3} > H_{2}O > NH_{2}^{-} > OH^{-}$$

D
$$H_{2}O > NH_{3} > OH^{-} > NH_{2}^{-}$$

Solution

Basic strength depends upon the electron donating capacity of the central atom, here amide is most basic due to presence of negative charge and two lone pair of electrons on nitrogen atom.

$$NH_{2}^{-} > OH^{-} > NH_{3} > H_{2}O$$
Question 3
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Which of the following statements is not correct?

A
Primary amines show intermolecular hydrogen bonding.

B
Secondary amines show intermolecular hydrogen bonding.

C
Tertiary amines show intermolecular hydrogen bonding.

D
Amines have lower boiling points as compared to those of alcohols and carboxylic acids of comparable molar masses.

Solution

For hydrogen bonding to occur, we need a polarized hydrogen atom (a hydrogen atom attached directly to $$F$$, $$O$$ or $$N $$ is called as a polarized hydrogen).
In a tertiary amine, there is no polarized hydrogen because all the hydrogens are replaced by alkyl or aryl groups.
So, tertiary amines cannot show hydrogen bonding.
Question 4
1 / -0

Which of the following undergoes diazotization?

A
$${ CH }_{ 3 }{ NH }_{ 2 }$$

B
$${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$$

C
$${ CH }_{ 3 }{ CONH }_{ 2 }$$

D
$${ CH }_{ 3 }N\left( { CH }_{ 3 } \right) _{ 2 }$$

Solution

Only primary aromatic amines undergo diazotization to give diazonium salts.
Reaction is as follows :

$$C_{6}H_{5}NH_{2} + NaNO_{2} \xrightarrow {HCl} C_{6}H_{5}N_{2}^{+}Cl^{-}$$

So, B is the correct answer because it is a primary aromatic amine.
Question 5
1 / -0

$${ CH }_{ 3 }{ CH }_{ 2 }Cl\xrightarrow { NaCN } X\xrightarrow { Pt/{ H }_{ 2 } } Y\xrightarrow [ anhydride ]{ Acetic } Z$$
In the above sequence, $$Z$$ is:

A
$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ NHCOCH }_{ 3 }$$

B
$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ NH }_{ 2 }$$

C
$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CONHCH }_{ 3 }$$

D
$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CONHCOCH }_{ 3 }$$

Solution

$$CH_{3}CH_{2}Cl + NaCN \rightarrow CH_{3}CH_{2}CN$$

So, $$X$$ is $$CH_{3}CH_{2}CN$$ and is formed by nucleophilic substitution reaction ($$S_N2$$)

Now, $$CH_{3}CH_{2}CN + H_{2}/Pt \rightarrow CH_{3}CH_{2}CH_{2}NH_{2}$$

So, $$Y$$ is $$CH_{3}CH_{2}CH_{2}NH_{2}$$ and is formed by reduction reaction. The cyanide group on reduction gives amino group.
$$CH_{3}CH_{2}CH_{2}NH_{2} + CH_{3}COOCOCH_{3} \rightarrow CH_{3}CH_{2}CH_{2}NHCOCH_{3}$$

So, $$Z$$ is $$CH_{3}CH_{2}CH_{2}NHCOCH_{3}$$ and is formed by addition-elimination reaction.
Question 6
1 / -0

The correct order of basicity of the given compounds is :

A
$$CH_{3} CH_{2} NH_{2} < HO(CH_{2})_{3}NH_{2}< HO(CH_{2})_{2}NH_{2}$$

B
$$CH_{3} CH_{2} NH_{2} < HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2}$$

C
$$CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2}$$

D
$$CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}< HO(CH_{2})_{2}NH_{2}$$

Solution

The correct order of basicity of the given compounds is as follows:

$$\displaystyle CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2} $$

$$OH$$ group is an electron withdrawing group and hence, it increases the partial positive charge on $$N$$ atom. Therefore, lone pair of $$N$$ cannot be easily donated and also, the basicity decreases.

As the number of carbon atoms between $$-OH$$ group and amino group increases, the electron withdrawing capacity of $$ -OH$$ group on amine $$N$$ decreases and basicity increases.
Question 7
1 / -0

Arrange the following in the correct order of their basic character in the aqueous phase.

$$(i)$$ $${ NH }_{ 3 }$$

$$(ii)$$ $${ RNH }_{ 2 }$$

$$(iii)$$ $${ R }_{ 2 }NH$$

$$(iv)$$ $${ R }_{ 3 }N$$

A
$$IV > III > II > I$$

B
$$III > IV > II > I$$

C
$$III > II > IV > I$$

D
$$I > II > III > IV$$
Question 8
1 / -0

Which of the following when treated with aniline will give benzene diazonium chloride as the major product?

A
$${ NaNO }_{ 2 }$$ and $$HCl$$ between $$0-{ 5 }^{ 0 }C$$

B
$${ HNO }_{ 3 }$$ and $$HCl$$ between $$0-{ 5 }^{ 0 }C$$

C
$${ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }$$ between $$0-{ 5 }^{ 0 }C$$

D
$${ NaNO }_{ 2 }$$ between $$0-{ 5 }^{ 0 }C$$

Solution

Benzene diazonium chloride is $$PhN_{2}^{+}Cl^{-}$$.

It is formed when aniline is treated with $$NaNO_{2}-HCl$$ at a temperature between 0-5$$^{\circ} C $$.

$$PhNH_{2} + NaNO_{2}+HCl \rightarrow PhN_{2}^{+}Cl^{-}$$

Benzene diazonium chloride is an important synthetic intermediate and is used in a number of reactions.

Option A is correct.
Question 9
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$${ CaC }_{ 2 }\xrightarrow { Hydrolysis } A\xrightarrow [ Cu\:tube ]{ Red\:hot } B\xrightarrow [ { 50-60 }^{ 0 }C ]{ { HNO }_{ 3 }-{ H }_{ 2 }{ SO }_{ 4 } } C\xrightarrow { Fe+HCl } D \xrightarrow { { NaNO }_{ 2 }+HCl,\:{ 0}^{ 0 }C } E$$

In the reaction sequence given above, $$E$$ is:

A
aniline

B
benzene diazonium chloride

C
phenyl osazone

D
benzoyl chloride

Solution

Hydrolysis of calcium carbide yields ethyne.
$$CaC_{2} + 2H_{2}O \rightarrow C_{2}H_{2} + Ca(OH)_{2}$$

Now, in the presence of red hot copper tube, $$3$$ molecules of ethyne polymerize to give benzene.
$$3C_{2}H_{2}$$ + red hot $$Cu$$ tube $$\rightarrow C_{6}H_{6}$$ (benzene)

In the presence of $$HNO_{3}-H_{2}SO_{4}$$, nitration of benzene occurs (electrophilic aromatic substitution) and nitrobenzene is formed.
$$C_{6}H_{6} + HNO_{3}-H_{2}SO_{4} \rightarrow C_{6}H_{5}NO_{2}$$ (nitrobenzene)

In the presence of $$Fe-HCl$$, nitrobenzene reduces to aniline.
$$C_{6}H_{5}NO_{2} + Fe-HCl \rightarrow C_{6}H_{5}NH_{2}$$ (aniline)

Aniline on reaction with $$NaNO_{2}-HCl$$ gives benzene diazonium chloride.
$$C_{6}H_{5}NH_{2} + NaNO_{2}-HCl \rightarrow C_{6}H_{5}N_{2}^{+}Cl^{-}$$

The complete reaction sequence is as follows:

$${ CaC }_{ 2 }\xrightarrow { Hydrolysis } C_2H_2\xrightarrow [ Cu\:tube ]{ Red\:hot } C_6H_6\xrightarrow [ { 50-60 }^{ 0 }C ]{ { HNO }_{ 3 }-{ H }_{ 2 }{ SO }_{ 4 } } C_6H_5NO_2\xrightarrow { Fe+HCl } C_6H_5NH_2$$

$$ \xrightarrow {NaNO_2 + HCl,\: 0^oC} C_6H_5N_2^+Cl^-$$

Therefore, $$E$$ is benzene diazonium chloride.
Question 10
1 / -0

Which of the following has the maximum value of $${ pk_b }$$?

A
$$\left( { CH }_{ 3 } \right) _{ 2 }NH$$

B
$$\left( { CH }_{ 3 }{ CH }_{ 2 } \right) _{ 2 }NH$$

C
$$C_6H_5-NH-C_6H_5$$

D
$$C_6H_5-NH-CH_3$$

Solution

In the compound mentioned in part C, the lone pair of nitrogen is in conjugation with both the phenyl rings.

So, they are greatly dispersed and are not available for donation.

As a result, the compound C is least basic.

Least basic means least $$K_{b}$$ value and least $$K_{b}$$ value means highest $$pK_{b}$$ value.

Option C is correct.

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Amines Test - 50

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Amines Test - 50

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Question 1

Allylmethylemine

$$2-$$ amino $$-4-$$ pentene

$$4-$$ aminopent $$-1-$$ ene

$$N-$$ methylprop $$-2-$$ en $$-1-$$ amine

Solution

Question 2

$$NH_{2}^{-} > OH^{-} > NH_{3} > H_{2}O$$

$$OH^{-} > NH_{2}^{-} > H_{2}O > NH_{3}$$

$$NH_{3} > H_{2}O > NH_{2}^{-} > OH^{-}$$

$$H_{2}O > NH_{3} > OH^{-} > NH_{2}^{-}$$

Solution

Question 3

Primary amines show intermolecular hydrogen bonding.

Secondary amines show intermolecular hydrogen bonding.

Tertiary amines show intermolecular hydrogen bonding.

Amines have lower boiling points as compared to those of alcohols and carboxylic acids of comparable molar masses.

Solution

Question 4

$${ CH }_{ 3 }{ NH }_{ 2 }$$

$${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$$

$${ CH }_{ 3 }{ CONH }_{ 2 }$$

$${ CH }_{ 3 }N\left( { CH }_{ 3 } \right) _{ 2 }$$

Solution

Question 5

$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ NHCOCH }_{ 3 }$$

$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ NH }_{ 2 }$$

$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CONHCH }_{ 3 }$$

$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CONHCOCH }_{ 3 }$$

Solution

Question 6

$$CH_{3} CH_{2} NH_{2} < HO(CH_{2})_{3}NH_{2}< HO(CH_{2})_{2}NH_{2}$$

$$CH_{3} CH_{2} NH_{2} < HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2}$$

$$CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}> HO(CH_{2})_{2}NH_{2}$$

$$CH_{3} CH_{2} NH_{2} > HO(CH_{2})_{3}NH_{2}< HO(CH_{2})_{2}NH_{2}$$

Solution

Question 7

$$IV > III > II > I$$

$$III > IV > II > I$$

$$III > II > IV > I$$

$$I > II > III > IV$$

Question 8

$${ NaNO }_{ 2 }$$ and $$HCl$$ between $$0-{ 5 }^{ 0 }C$$

$${ HNO }_{ 3 }$$ and $$HCl$$ between $$0-{ 5 }^{ 0 }C$$

$${ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }$$ between $$0-{ 5 }^{ 0 }C$$

$${ NaNO }_{ 2 }$$ between $$0-{ 5 }^{ 0 }C$$

Solution

Question 9

aniline

benzene diazonium chloride

phenyl osazone

benzoyl chloride

Solution

Question 10

$$\left( { CH }_{ 3 } \right) _{ 2 }NH$$

$$\left( { CH }_{ 3 }{ CH }_{ 2 } \right) _{ 2 }NH$$

$$C_6H_5-NH-C_6H_5$$

$$C_6H_5-NH-CH_3$$

Solution

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